User:Egm6321.f10.team3.cook/HW7

=Problem 8: Find the Binomial expansion of $$ { \left(1-x \right) }^{- \frac{1}{2}} $$ =

Given
From [|Mtg 40-4] of the lecture notes we have the Binomial Theorem generalized for real numbers.


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$$ { \left(x+y\right) }^{r}= \sum_{k=0}^{\infty} \binom{r}{k}{x}^{r-k}{y}^{k}$$ $$
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 * $$\displaystyle (Eq. 8.1)
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Where the Binomial coefficient


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$$ \binom{r}{k}= \frac{r \left( r-1 \right) \left( r-2 \right).... \left( r-k+1 \right)}{k!} $$ $$
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 * $$\displaystyle (Eq. 8.2)
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Note: By the [| Multiplicitve Formula] $$\binom{r}{k} = \frac{{2}^{r}}{k!}$$. This allows the Binomial coefficient to be generalized to include real and complex numbers.

Find
Use the General Binomial Theorem to obtain (6)&(7) from [|Mtg 40-3] of the lecture notes:


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{\left(1-x\right) }^{- \frac{1}{2}}= \sum_{i=0}^{\infty} {\alpha }_{i}{x}^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8.3)
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{ \alpha }_{i} = \frac{1*3*5..... \left(2i-1 \right)}{2*4*6.... \left( 2i\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8.4)
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Solution
Apply the General Binomial Theorem for the parameters $$\displaystyle r= -\frac{1}{2} ; y = 1 ; x = -x$$


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{ \left(1-x\right) }^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{ -\frac{1}{2}}{k} {x}^{- \frac{1}{2}-k} {1}^{k}$$ $$ Using the hint provided in the lecture notes we will assume that $$ \left| x\right|  < 1 $$ to obtain
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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{ \left(1-x\right) }^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{ -\frac{1}{2}}{k} {x}^{k} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8.5)
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Now we evaluate the binomial coefficient using $$\displaystyle (Eq. 8.2)$$ we have


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\binom{ -\frac{1}{2}}{k}= \frac{-\frac{1}{2} \left( -\frac{1}{2}-1 \right) \left( -\frac{1}{2}-2 \right).... \left( -\frac{1}{2}-k+1 \right)}{k!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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Now we pull out the common factor of $$ \frac{1}{2} $$ and distribute through the negative we obtain


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\binom{ -\frac{1}{2}}{k}= \frac{1}{2}\frac{1 \left( 1+2 \right) \left( 1+4 \right).... \left( 1+2k - 2 \right)}{k!}=\frac{\left( 1 \right) \left( 3\right) \left( 5  \right).... \left( 2k - 1 \right)}{2k!} = \frac{\left( 1 \right) \left( 3\right) \left( 5  \right).... \left( 2k - 1 \right)}{\left( 2 \right) \left( 4  \right) \left( 6 \right).... \left( 2k \right) } $$ $$ We can see that $$ k=i$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8.6)
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\binom{ -\frac{1}{2}}{k}=  { \alpha }_{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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Changing the index from k to i we have:
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{\left(1-x\right) }^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{ -\frac{1}{2}}{k} {x}^{k} = \sum_{i=0}^{\infty} {\alpha }_{i}{x}^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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Thus (6) & (7) can be obtained from the General Binomial Theorem.

Contributing Members
Solved and Posted by --Egm6321.f10.team3.cook 13:28, 7 December 2010 (UTC)